Absolute Value and the Triangle Inequality

The triangle inequality for $|\cdot|$ is the seed of the general triangle inequality axiom in every metric space, used throughout topology, functional analysis, and geometry (norms on vector spaces, $L^p$ spaces, the Cauchy–Schwarz-derived triangle inequality in Euclidean space). In numerical analysis and scientific computing, absolute-value bounds control rounding and truncation error propagation: an error bound on a computed sum is obtained exactly via finite subadditivity. In statistics and probability, $|X - \mu|$ measures deviation from a mean, and Chebyshev-type and concentration inequalities are triangle-inequality arguments dressed in probabilistic language. In cryptography and coding theory, Hamming and related distances satisfy triangle-inequality axioms modeled directly on this one, enabling error-correcting code design. In physics and engineering, absolute error and tolerance analysis (e.g., propagating measurement uncertainty through a sum of readings) is literally the computation performed in Example 2 above. In computer science, floating-point stability proofs and algorithm error analysis (e.g., in numerical linear algebra) chain triangle-inequality steps to bound accumulated error over many operations.

Ask the tutor: Can you show me how the triangle inequality is used to prove that the limit of a product of sequences equals the product of the limits?

Think of the real line as a physical ruler, with $0$ as a fixed reference point. The number $|x|$ answers the question "how far is $x$ from $0$, ignoring direction?" A positive number is already its own distance from the origin; a negative number's distance from the origin is obtained by flipping its sign, since distance cannot be negative. This is exactly what the case-split definition encodes: it is a rule for stripping away the sign while preserving magnitude. The triangle inequality generalizes this from distances-to-the-origin to distances between arbitrary points. If you travel from $a$ to $c$, you cannot do better than going in a straight line; any detour through an intermediate point $b$ is at least as long. Algebraically, $|x+y| \le |x| + |y|$ is this principle applied with $a = 0$, $b = x$, $c = x+y$ (so the two "legs" have lengths $|x|$ and $|y|$, and the direct path has length $|x+y|$). Equality happens precisely when there is no wasted detour — when $x$ and $y$ point in the same direction (have the same sign, or one is zero).

lemma[Sign bound] For every $x \in \mathbb{R}$, $-|x| \le x \le |x|$.

::: Proof of lemma. If $x \ge 0$: $|x| = x$, so $x \le |x|$ holds with equality, and $-|x| = -x \le 0 \le x$ (since $x\ge 0$). If $x < 0$: $|x| = -x$, so $x \le |x|$ becomes $x \le -x$, true since $x < 0 < -x$; and $-|x| = x \le x$, an equality. In both cases $-|x| \le x \le |x|$. $\blacksquare$ Now apply the lemma to $x$ and to $y$ and add the two chains of inequalities (addition preserves order-compatible inequalities termwise, an order-axiom consequence):

Set $a = |x|+|y| \ge 0$ (a sum of nonnegatives, by part 1). We have shown $-a \le x+y \le a$. By part (4) applied with this $a$, this is exactly the statement $|x+y| \le a = |x|+|y|$. This proves the inequality. Equality case. We claim equality $|x+y| = |x|+|y|$ holds iff $xy \ge 0$. ($\Leftarrow$) Suppose $xy \ge 0$. By trichotomy on $x$ and $y$ combined with $xy\ge 0$, either both $x,y \ge 0$ or both $x, y \le 0$ (if one were strictly positive and the other strictly negative, their product would be negative by the order axioms, contradicting $xy \ge 0$). If both $\ge 0$: $x+y \ge 0$, so $|x+y| = x+y = |x|+|y|$. If both $\le 0$: $x + y \le 0$, so $|x+y| = -(x+y) = (-x)+(-y) = |x|+|y|$. ($\Rightarrow$) Suppose $xy < 0$ (i.e. not $xy \ge 0$); we show equality fails. Then one of $x,y$ is strictly positive and the other strictly negative; say $x>0>y$ (the other case is symmetric). Then $|x|=x$, $|y|=-y$, so $|x|+|y| = x-y$. We compute $|x+y|$: since we cannot determine the sign of $x+y$ in general, use instead the strict version of the lemma's second inequality applied to $y$: because $y<0$, we have $-|y| = y < |y|$ is not what we need — instead argue directly. Since $y < 0 < x$, we have $y < |y|$ actually reduces to $y< -y$ which is just $y<0$, true but not directly useful; instead compare $x+y$ against $x - y = |x|+|y|$ directly: $x - y - (x+y) = -2y > 0$ since $y<0$, so $x+y < x-y = |x|+|y|$. Also $-(x+y) < x - y$ is checked similarly: $x-y-(-(x+y)) = x - y + x+y = 2x > 0$ since $x > 0$, so $-(x+y) < x-y$ too. Hence both $x+y$ and $-(x+y)$ are strictly less than $|x|+|y|$, and since $|x+y| \in \{x+y, -(x+y)\}$, we conclude $|x+y| < |x|+|y|$, strict inequality. This shows equality forces $xy \ge 0$, completing the biconditional.

(6) Reverse triangle inequality. Write $x = (x-y) + y$. By the triangle inequality (part 5) applied to the two numbers $x-y$ and $y$:

Rearranging (an order-axiom consequence of adding $-|y|$ to both sides):

Symmetrically, writing $y = (y-x)+x$ and using $|-(x-y)| = |x-y|$ (part 2):

i.e. $-(|x|-|y|) \le |x-y|$, equivalently $|x|-|y| \ge -|x-y|$. Combining the two displayed inequalities,

By part (4) with $a = |x-y| \ge 0$, this is exactly $\big||x|-|y|\big| \le |x-y|$.

Metric properties of $d$. Positivity and the equality case follow immediately from part (1) applied to $x-y$. Symmetry $d(x,y)=d(y,x)$ follows from $|x-y| = |-(y-x)| = |y-x|$ by part (2). The triangle inequality for $d$ follows from part (5): writing $x - z = (x-y)+(y-z)$,

$\blacksquare$

Example 1 (Solving an inequality with the bounding property). Solve $|3x - 7| \le 5$ for $x \in \mathbb{R}$, and express the solution as an interval. By part (4) of the theorem with $a = 5 \ge 0$:

Add $7$ to all three parts (order-preserving under addition):

Divide by $3 > 0$ (order-preserving under multiplication by a positive number):

So the solution set is the closed interval $\left[\tfrac{2}{3}, 4\right]$. As a check, test an endpoint and an interior point: at $x=4$, $|3(4)-7| = |5| = 5 \le 5$ ✓; at $x = \tfrac23$, $|3(\tfrac23)-7| = |2-7| = 5 \le 5$ ✓; at $x=1$ (interior), $|3-7|=4 \le 5$ ✓. Test a point outside, $x=5$: $|15-7|=8 > 5$, correctly excluded.

Example 2 (Using the triangle inequality to bound a quantity, with asymmetric numbers). Suppose $a$ satisfies $|a - 3| < 0.2$ and $b$ satisfies $|b + 8| < 0.5$. Bound $|a + b + 5|$. We want to relate $a+b+5$ to the quantities we control, $a - 3$ and $b+8$. Observe the algebraic identity

which we verify by expanding the right side: $(a-3)+(b+8) = a+b+5$. ✓ (This is the key decomposition step — chosen so that each piece is exactly the quantity bounded by a hypothesis.) By the triangle inequality (part 5),

So $|a+b+5| < 0.7$. To see the bound is essentially tight (cannot be improved much), take $a = 3.19$ (so $|a-3| = 0.19 < 0.2$) and $b = -7.51$ (so $|b+8| = 0.49 < 0.5$). Then $a+b+5 = 3.19 - 7.51 + 5 = 0.68$, and indeed $0.68 < 0.7$, approaching the bound as $a,b$ approach the endpoints $3.2$ and $-7.5$. We can also illustrate the reverse triangle inequality numerically here: with $x = a = 3.19$ and $y=-7.51$,

and indeed $4.32 \le 10.7$ ✓, consistent with part (6).